Problem
In a study of cell phone usage and brain hemispheric dominance, an Internet survey was e-mailed to 6957 subjects randomly selected from an online group involved with ears. There were 1334 surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than $20 \%$. Use the P-value method and use the normal distribution as an approximation to the binomial distribution. Identify the null hypothesis and alternative hypothesis. A. $\mathrm{H}_{0}: \mathrm{p}>0.2$ $H_{1}: p=0.2$ C. $\mathrm{H}_{0}: \mathrm{p}<0.2$ $H_{1}: p=0.2$ E. $\mathrm{H}_{0}: \mathrm{p} \neq 0.2$ $\mathrm{H}_{1}: \mathrm{p}=0.2$ B. \[ \begin{array}{l} H_{0}: p=0.2 \\ H_{1}: p<0.2 \end{array} \] D. \[ \begin{array}{l} H_{0}: p=0.2 \\ H_{1}: p \neq 0.2 \end{array} \] F. \[ \begin{array}{l} H_{0}: p=0.2 \\ H_{1}: p>0.2 \end{array} \] The test statistic is $\mathrm{z}=-1.72$. (Round to two decimal places as needed.) The P-value is $\square$. (Round to three decimal places as needed.)
Solution
Step 1 ：Identify the null hypothesis and alternative hypothesis. The null hypothesis is \(H_{0}: p=0.2\) and the alternative hypothesis is \(H_{1}: p<0.2\).
Step 2 ：Calculate the test statistic, which is \(z=-1.72\).
Step 3 ：Calculate the P-value using the cumulative distribution function. The P-value is approximately 0.043.
Step 4 ：Final Answer: The null hypothesis and alternative hypothesis are: \[ \begin{array}{l} H_{0}: p=0.2 \ H_{1}: p<0.2 \end{array} \] The P-value is \(\boxed{0.043}\).
